Differentiating When y Isn’t Isolated: Implicit and Higher-Order Derivatives

Implicit relationship

An equation connecting $x$ and $y$ where $y$ is not isolated (e.g., $x^2 + y^2 = 25$), describing a curve even if it is not a function $y = f(x)$ globally.

Implicit differentiation

A technique for finding dy/dx by differentiating both sides of an equation with respect to x without first solving for y.

Treat y as a function of x

The implicit-differentiation mindset that y depends on x along the curve, so derivatives of expressions involving y require the Chain Rule.

Chain Rule “receipt” factor

When differentiating an expression involving y with respect to x, you multiply by dy/dx to account for y changing as x changes.

Derivative of $y^2$ (implicit)

$\frac{d}{dx}(y^2) = 2y \cdot \frac{dy}{dx}$.

Derivative of $y^3$ (implicit)

$\frac{d}{dx}(y^3) = 3y^2 \cdot \frac{dy}{dx}$.

Derivative of sin(y) (implicit)

$\frac{d}{dx}(\sin(y)) = \cos(y)\bullet\frac{dy}{dx}$.

Derivative of cos(y) (implicit)

$\frac{d}{dx}(\cos(y)) = -\sin(y) \bullet \frac{dy}{dx}$.

dy/dx notation

The derivative of y with respect to x; in implicit problems it represents the slope of the tangent line in terms of x and y.

y′ notation

An alternative notation for dy/dx; it means the same first derivative.

d/dx[F] notation

Operator notation meaning “differentiate F with respect to x,” useful when differentiating both sides of an equation.

Implicit differentiation algorithm

Differentiate both sides w.r.t. x, apply Chain Rule to y-terms, collect all dy/dx terms on one side, factor out dy/dx, and solve.

Product Rule for xy (implicit setting)

$\frac{d}{dx}(xy) = x\bullet\frac{dy}{dx} + y$, because $y$ depends on $x$.

Circle slope via implicit differentiation

For $x^2 + y^2 = 25$, implicit differentiation gives $\frac{dy}{dx} = -\frac{x}{y}$.

Tangent line (point-slope form)

A line using slope $m$ and point $(x_1,y_1)$: $y - y_1 = m(x - x_1)$, often with $m$ found from $\frac{dy}{dx}$.

Tangent line at $(3,4)$ on $x^2 + y^2 = 25$

Since $\frac{dy}{dx} = -\frac{x}{y}$, the slope at $(3,4)$ is $-\frac{3}{4}$, so $y - 4 = -\frac{3}{4}(x - 3)$.

Example: implicit derivative of $xy + \sin(y) = x^2$

Differentiating gives $x\frac{dy}{dx} + y + \cos(y)\frac{dy}{dx} = 2x$, so $\frac{dy}{dx} = \frac{2x - y}{x + \cos(y)}$.

Common Chain Rule mistake (sin(y))

Incorrect: $\frac{d}{dx}(\sin(y)) = \cos(y)$. Correct: $\cos(y)\bullet\frac{dy}{dx}$.

Horizontal tangent condition (implicit)

If $\frac{dy}{dx} = \frac{N(x,y)}{D(x,y)}$, a horizontal tangent occurs when $N(x,y) = 0$ and $D(x,y) \neq 0$, at a point on the original curve.

Vertical tangent condition (implicit)

If $\frac{dy}{dx} = \frac{N(x,y)}{D(x,y)}$, a vertical tangent occurs when $D(x,y) = 0$ and $N(x,y) \neq 0$, at a point on the original curve.

Verification step for tangent candidates

After setting numerator or denominator conditions for horizontal/vertical tangents, you must check the candidate points satisfy the original equation.

Higher-order derivative (implicit context)

Any derivative beyond the first (e.g., y'' = d^2y/dx^2); found by differentiating y′ while remembering y and y′ depend on x.

Concavity via second derivative

If y'' > 0 the curve is concave up (slopes increasing); if y'' < 0 the curve is concave down (slopes decreasing).

Second derivative of the circle $x^2 + y^2 = 25$

Starting from $y' = -\frac{x}{y}$, one form is $y'' = -\frac{y^2 + x^2}{y^3}$, which simplifies using $x^2 + y^2 = 25$ to $y'' = -\frac{25}{y^3}$.

Product Rule for x·y′ when finding y''

$\frac{d}{dx}(x \times \frac{dy}{dx}) = x \times \frac{d^2y}{dx^2} + \frac{dy}{dx}$, because both $x$ and $y'$ depend on $x$.